The physical basis of material properties like Young’s modulus can be understood by examining materials on the atomic scale. There are two main things that influence the value of the modulus:
1.) The atomic microstructure
2.) The interatomic bonds.
Different values are obtained for the elastic modulus depending upon the crystallographic direction in which we measure E. This directional variation in properties is known as anisotrophy. For example, the elastic modulus for a single crystal of iron varies between 41,000 ksi and 19,000 ksi, depending upon the direction of measurement. Tabulated values of E are usually average values taken from polycrystalline materials with a random orientation of the individual grains.
ATOMIC MICROSTRUCTURE
All solid materials may be classified as either crystalline or amorphous based upon the way in which the atoms arrange themselves. Crystalline materials are characterized by long range order. This means that the atoms arrange themselves into regular, repeating, three-dimensional patterns. The crystals formed by these rather large groups of atoms are called grains. An example of a crystalline structure would be the zinc coating on a galvanized steel sheet. Amorphous solids do not possess any long range order, although they may have short range order. Glass is a good example of this type of material.
The fact that crystalline solids have long range order means that the atom or group of atoms that make up the basic unit of the material must have identical surroundings. If we model the atoms as hard spheres, then we can think of packing them together in a plane as though we were racking a set of billiard balls for a game of pool. The balls are arranged so that they take up the least amount of space. In this two-dimensional example this type of plane is called a close-packed plane, and the directions along which the balls touch are called close-packed directions. We could extend this pattern by adding balls until it completely covers the pool table. The important thing to notice is that the balls are arranged in a regular repeating two dimensional pattern.
Now suppose that we start adding balls on top of the first plane that we already arranged. How we position the second plane of atoms is important, because it will determine the type of three dimensional structure that will be produced. The depressions that are formed in the first plane of atoms where three atoms touch are ideal locations for the atoms in the second layer to sit. By dropping atoms into these convenient “seats” we can build a second close packed plane on top of the first one. By adding more planes on top of the previous ones in this way, we find that we can produce a three dimensional structure where the atoms take up the least amount of space. This is an example of a close packed structure. FCC is one microstructure that can be formed using this type of construction.
ATOMIC BONDS
The strength of an interatomic bond depends upon the forces that exist between the bonding atoms. From a theoretical standpoint we can determine the force F between two atoms for any separation distance r from the relationship
F = dU/dr
where U® is the interatomic potential function. F is zero at the equilibrium point r = ro. Suppose that the atoms are pulled apart to a separation distance r which displaces them from the equilibrium separation distance ro by an amount (r - ro). In this case a resisting force appears. For small displacements from equilibrium (r - ro) the resisting force is proportional to the displacement for all materials in both tension and compression.
The stiffness S of the resulting bond is given by
S = dF/dr = d^2U/dr^2
If the bond is not stretched too far, S is approximately constant and is given by
So = (d^2U/dr^2) evaluated at r = ro
So the bond behaves in a linear elastic manner. This is the physical origin of Hooke’s Law. A narrow, steep potential well corresponds to a stiff material with a high modulus. A broad, shallow potential well represents a material with a low modulus. You can think of So as the “spring constant” of this tiny atomic spring. So the force between a pair of atoms stretched apart to a distance r, where r is slightly greater than (or less than) ro, is
F = So(r-ro)
Now imagine a solid held together by these linear springs, joining two adjacent planes of atoms together. The number of bonds that are formed between the atoms in these two adjacent planes will directly impact the mechanical response that the material has to an applied stress. The greater the number of bonds that are formed per atom (which is a direct result of the atomic microstructure), the more resistant the material is to deforming under load. The stronger the forces that exist between the atoms (which is a direct result of the bonding), the more resistant the material is to deformation. So each of these parameters influences the response of the material to an applied stress. For simplicity, imagine that we are dealing with a material that has ionic bonds, with a simple cubic crystal structure. If our adjacent planes are stretched apart so that they are displaced from their equilibrium separation distance by an amount (r-ro), then the total force per unit area, defined as the stress, is given by
Stress = NSo(r-ro)
where N is the number of bonds per unit area. If we draw a simple cubic structure, we find that the atoms are spaced a distance ro apart, with each atom at the corner of a cube. So the average area per atom is equal to ro^2. It follows that N = 1/ro^2. We can convert displacement from equilibrium (r-ro) into strain by dividing by the initial separation distance ro to obtain
strain = (r-ro)/ro
so that
stress = (So/ro)*strain
Young’s modulus is therefore
E = stress/strain = So/ro
If you know the interatomic potential function, then you can calculate a theoretical value for the modulus based upon this simple model. Keep in mind that the crystal structure that was assumed is simple cubic, which most materials do not possess. And we have also ignored the effects of secondary, or long range bonding as well. Still, the results that are obtained from this calculation are surprisingly accurate when compared to actual measured values of real materials. For the ionic bond, the interatomic potential function can be expressed in the form
U(r) = constant - q^2/[4*pi*epsilon]+ B/r^n
where q is the electron charge, pi=3.1412, epsilon is the permittivity of free space, B is a constant for a particular material, and n is an integer. Differentiating this expression once, setting r=ro, and setting the resulting expression equal to zero we can solve for B. We find
B = [(q^2)*ro^(n-1)]/[4*pi*n*epsilon]
So the stiffness of the bond is given by
So = [alpha*q^2]/[4*pi*epsilon*ro^3]
where alpha = n-1. Consider the case of NaCl. The coulombic attraction in this ionic bond is a long range interaction that varies as 1/r. Because of this, an Na+ ion not only interacts attractively with its neighboring Cl- ions, but it also interacts repulsively with its slightly more distant Na+ neighbors. To calculate So properly, we must sum over all of these bonds, taking both attractions and repulsions into account. Doing this, we find that the result is the above expression for So with a value of alpha=0.58. Substituting in values for the physical constants as well as using a value for the atomic spacing of 2.5 angstroms, we find that So = 8.54 N/m. In the case of NaCl this results in a value for E of 34.2 GPa. The measured value of E for NaCl is 39.96 GPa. So the calculated result is within 15% of the actual value.
And, thank you to @Maui for this summary explanation.